Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 11x}{x + 8} = \dfrac{-23x - 32}{x + 8}$
Solution: Multiply both sides by $x + 8$ $ \dfrac{x^2 - 11x}{x + 8} (x + 8) = \dfrac{-23x - 32}{x + 8} (x + 8)$ $ x^2 - 11x = -23x - 32$ Subtract $-23x - 32$ from both sides: $ x^2 - 11x - (-23x - 32) = -23x - 32 - (-23x - 32)$ $ x^2 - 11x + 23x + 32 = 0$ $ x^2 + 12x + 32 = 0$ Factor the expression: $ (x + 4)(x + 8) = 0$ Therefore $x = -4$ or $x = -8$ At $x = -8$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -8$, it is an extraneous solution.